search for: left_join

...la[, PorcAc := cumsum(Porc)] # Regiones tabla.1 <- datos[, .N, by = .(etiqueta = x, Region)] tabla.1[, Porc := round(N/sum(N)*100,1), by =.(Region)] tabla.1 <- spread(tabla.1[, .(etiqueta, Region, Porc)], key = Region, value = Porc) tabla.reg <- left_join(tabla, tabla.1) # Depdendencia tabla.1 <- datos[, .N, by = .(etiqueta = x, Dependencia = B2a)] tabla.1[, Porc := round(N/sum(N)*100,1), by =.(Dependencia)] # % por Dependencia tabla.1 <- na.omit(tabla.1) tabla.1 <- spread(tabla.1[, .(etique...

...[, PorcAc := cumsum(Porc)]           # Regiones           tabla.1 <- datos[, .N, by = .(etiqueta = x, Region)]            tabla.1[, Porc := round(N/sum(N)*100,1), by =.(Region)]            tabla.1 <- spread(tabla.1[, .(etiqueta, Region, Porc)], key = Region, value = Porc)      tabla.reg <- left_join(tabla, tabla.1)           # Depdendencia           tabla.1 <- datos[, .N, by = .(etiqueta = x, Dependencia = B2a)]            tabla.1[, Porc := round(N/sum(N)*100,1), by =.(Dependencia)] # % por Dependencia           tabla.1 <- na.omit(tabla.1)           tabla.1 <- spread(tabla.1[, .(etiqu...

...for your help. Ivan, your solution worked perfectly. I didn't really understand how to do string processing on a vector of strings, and your solution demonstrated it for me. I modified it to work with the tidyverses' stringr library in this way: bg3_race_sum <- bg3_race %>% left_join(pl_vars, by=c("variable" = "name")) %>% group_by(variable) %>% summarize(count = sum(value)) %>% left_join(pl_vars, by=c("variable" = "name")) %>% filter(count > 0) %>% .$label %>% str_replace("^ !!&quot...

...n it i can use it in future by doing some Lil bit manipulation. Thanks data_help <- data_help %>% mutate(Purchase_ID = 1:n()) %>% group_by(Purchase_ID) %>% do(split_items(.)) cat_help %>% gather("Foo", "Item") %>% filter(!is.na(Item)) %>% left_join(data_help, by = "Item") %>% group_by(Foo, Purchase_ID) %>% summarise(Item = paste(Item, collapse = ", ")) %>% spread(key = "Foo", value = "Item") On 31 August 2017 at 13:17, Ulrik Stervbo <ulrik.stervbo at gmail.com> wrote: > Hi Hema...

...=30)) %>% tbl_df() # instead of doing this separately for each factor ... df2 <- df1 %>% group_by(group) %>% dplyr::count(var1, wt=wt) %>% mutate(prop1=n/sum(n)) df3 <- df1 %>% group_by(group) %>% dplyr::count(var2, wt=wt) %>% mutate(prop2=n/sum(n)) %>% left_join(df2, by='group') # I would like to do something like the following (which does of course not work): my_fun <- function(x,wt){ freq1 <- dplyr::count(x, wt=wt) prop1 <- freq1 / sum(freq1) return(prop) } df1 %>% group_by(group) %>% summarise_all(.funs=my_fun(.), .var...

Hola a todos, Quisiera juntar las informacion de dos data.frames con una union de columnas un tanto especial. La informacion que tengo son datos de captura-recaptura de diferentes individuos, por ejemplo en una base de datos tengo:ID <- c(1,2,3,4)Fate_1 <- c(2,2,2,2)Fate_2 <- c(0,0,0,NA)Fate_3 <- c(0, NA, NA, NA) y en otra base de datos tengo:ID <- c(1,2,3)Fate_1 <- c(0, 0, 0,

> On 7 Oct 2019, at 17:04, Tierney, Luke <luke-tierney at uiowa.edu> wrote: > > Think about what happens if an > argument in a pipe stage contains a pipe. (Not completely > unreasonable, e.g. for a left_join). It should work exactly as it does in a local environment. ``` `%foo%` <- function(x, y) { env <- parent.frame() # Use `:=` to avoid partial matching on .env/.frame rlang::scoped_bindings(. := x, .env = env) eval(substitute(y), env) } "A" %foo% { print(.) "B&q...

Buenos díasTengo esta pequeña matriz con tres columnas. Querría saber como puedo filtrar el mes en el que todos los campos de la columna RESULTADO son NA  (en este caso sería unicamente el mes 4). Gracias  | MES | VARIABLE | RESULTADO | | 1 | A | SI | | 1 | B | SI | | 1 | C | NO | | 2 | A | NA | | 2 | B | SI | | 2 | C | SI | | 3 | A | NO | | 3 | B | NO | | 3 | C | NO | | 4 | A | NA | | 4 | B | NA

...ames = FALSE) data_frame(Item = x, Purchase_ID = items$Purchase_ID) } data_help <- data_help %>% mutate(Purchase_ID = 1:n()) %>% group_by(Purchase_ID) %>% do(split_items(.)) cat_help %>% gather("Foo", "Item") %>% filter(!is.na(Item)) %>% left_join(data_help, by = "Item") %>% group_by(Foo, Purchase_ID) %>% summarise(Item = paste(Item, collapse = ", ")) %>% spread(key = "Foo", value = "Item") HTH Ulrik On Wed, 30 Aug 2017 at 13:22 Hemant Sain <hemantsain55 at gmail.com> wrote: &gt...

...>> table_1 <- gather(table_1, "Foo", "Item") %>% >> filter(!is.na(Item)) >> >> table_1 <- separate(table_1, col = "Item", into = c("Quantity", "Item"), >> sep = " ") >> >> table_3 <- left_join(table_1, table_2, by = "Item") %>% >> mutate(Item = paste(Quantity, Item)) %>% >> select(-Quantity) >> >> table_3 %>% >> group_by(Foo, Category) %>% >> summarise(Item = paste(Item, collapse = ", ")) %>% >> spread...

Buenas tardes, Estoy precisando generar una nueva variable que contenga el mes en tres letras, por ejemplo: ENE , FEB, MAR , ABR y así sucesivamente a partir de los valores que ahora tengo en el Dataset, que son 1, 2, 3, 4 y así sucesivamente. Entiendo que sería con mutate, pero consulto acerca del comando completo.. Gracias, Jesús _________________ *Jesús MARTÍN FRADE * Skype:

...mes(kids_province_data)[2] <- "Number" # sum the data by province kids_province_sums <- aggregate(.~Province, data = kids_province_data, sum) # join the data to the map names(tract)[6] <- "region" names(kids_province_sums)[1] <- "region" by_province <- left_join(tract, kids_province_sums) # create the data map kids_map <- ggplot(by_province, aes(map_id = region)) + #plots the map in by_province separating by region geom_map(aes(fill = Number), #generates aestheticsa for the plot ma...

Buenas tardes, quedarme con las muestras de una BD (data) que están presentes en otra (datax), cuando se tiene una variable que nunca se repite (Key) es fácil: data <- subset(data,data$Key %in% datax$Key). Mi problema es cuando la exclusividad viene dada por dos variables. P.e., las coordenadas de un mapa: lon y lat. ¿Como puedo quedarme con las muestras de una df cuya lon y lat son iguales a

On Mon, 7 Oct 2019, Lionel Henry wrote: > > >> On 7 Oct 2019, at 17:04, Tierney, Luke <luke-tierney at uiowa.edu> wrote: >> >> Think about what happens if an >> argument in a pipe stage contains a pipe. (Not completely >> unreasonable, e.g. for a left_join). > > It should work exactly as it does in a local environment. > > ``` > `%foo%` <- function(x, y) { > env <- parent.frame() > > # Use `:=` to avoid partial matching on .env/.frame > rlang::scoped_bindings(. := x, .env = env) > > eval(substitute(y), env)...

I think a simple reproducible example ("reprex") may be necessary for you to get a useful reply. Questions with vague specifications such as yours often result in going round and round with attempts to clarify what you mean without a satisfactory answer. Clarification at the outset with a reprex may save you and others a lot of frustration. Cheers, Bert On Sat, Nov 4, 2023 at 1:41?AM

Buenas, Tengo dos data.frames de la siguiente manera library(data.table) id<-c("a1","a2","a3","a4") id2<-c("a2","a3","a1","a4") y<-c(1,2,3,4) z<-c(3,5,6,7) k<-c(1,3,8,7) df1<-data.table(id,y,z) id<-c("a2","a3","a1","a4") df2<-data.table(id,x,y) Quiero

...in `on.exit()` (or through > unwind-protection if implemented in C). > Sorry to be blunt but adding/removing a variable from a caller's environment is a horrible design idea. Think about what happens if an argument in a pipe stage contains a pipe. (Not completely unreasonable, e.g. for a left_join). We already have such a design lurking in (at least) one place in base code and it keeps biting. It's pretty high on my list to be expunged. If a variable is to be used it needs to be in its own scope/environment. There is another option, which is to rewrite the pipe as a nested call and eva...

Hi Gabe, > There is another way the pipe could go into base R that could not be > done in package space and has the potential to mitigate some pretty > serious downsides to the pipes relating to debugging I assume you're thinking about the large stack trace of the magrittr pipe? You don't need a parser transformation to solve this problem though, the pipe could be implemented as

Hi Marna, This is a condition that the function cannot handle. It would be possible to reformat the result based on the time intervals, but the stretch_df function doesn't try to interpret the values, just stretches them out to a wide format. Jim On Wed, May 2, 2018 at 9:16 AM, Marna Wagley <marna.wagley at gmail.com> wrote: > Hi Jim, > The data set is correct. I took two

Hey PIKAL, It's not a homework neithe that is the real dataset i have signer NDA for my company so that i can share the original data file, Actually I'm working on a market basket analysis task but not able to convert my existing data table to appropriate format so that i can apply Apriori algorithm using R, and this is very important me to get it done because I'm an intern and if i